It turns out after researching the integrand for a bit that it has an \(extremely\) nice McLaurin series. I don't have the energy to type out the entire derivation, but you can check it's:
\[ \frac{x}{x^3+1}=x-x^4+x^7-x^{10}+...=x\sum_{n=0}^\infty(-1)^nx^{3n} \]
Throwing in this in the integral and doing a partial integration we get:
\[ \int_0^1{\frac{x}{x^3+1}\, dx}=\int_0^1{x\sum_{n=0}^\infty(-1)^nx^{3n}\,dx} \\
=\sum_{n=0}^\infty \frac{(-1)^n}{3n+1}-\sum_{n=0}^\infty(-1)^n\int_0^1{x^{3n}\,dx} \\
=\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}\left(1-\frac{1}{3n+2}\right) \\
=\sum_{n=0}^\infty\frac{(-1)^n}{3n+2} \]
Now that sum is a bit harder to calculate, it involves zeta-functions in combination with some subtitutions, but at least it comes out the same as wolf's answer :p

Having never taken complex analysis I really don't want to pretend that I understand much about integrating in the complex plane, though of course I understand the concept of writing complex numbers in them and roots of unity... basically.